Stoichiometry, Chemical Reactions, Chemical Thermodynamics, Chemical Kinetics

Chemical Equation ➼A condensed shorthand statement which expresses a chemical reaction in terms of formulas and symbols. ➼Indicates the reactants and the product/s of a chemical reaction ➼A balanced equation gives a quantitative relationship of the reactants and the products.

Chemical Reactions

❶Combination ❷Decomposition ❸Displacement ❹Replacement

Combination ➼simplest conceivable reaction ➼a reaction in which 2 or more substances unite to form a single compound ➼ex. Fe + S → FeS H2O + SO3 → H2SO4 H2O + Na2O → 2NaOH CaO + SiO2 → CaSiO3

Decomposition ➼reverse of combination reaction ➼a reaction in which a substances yields two or more products. ➼breaking up of binary compounds into its elements 2HgO → 2Hg + O2 ➼ex. 2KClO3 → 2 KCl + 3O2 CaCO3 → CaO + CO2

Displacement ➼a reaction in which compounds combine freeing one of its constituent element ➼ex. Zn + 2HCl → ZnCl2 + H2 Fe + CuSO4 → FeSO4 + Cu Cl2 + NaBr → 2NaCl + Br2 CaCO3 + SiO2 → CaSiO3 + CO2

Replacement ➼a reaction in which there is an interchange of elements (or radicals) between two compound ➼ex. AgF + NaCl → AgCl + NaF AgNO3 + NH4Cl → AgCl + NH4NO3 3Ca(NO3)2 + 2Na3PO4 → Ca3(PO4)2 + 6NaNO3 BaCl2 + NaSO4 → BaSO4 + 2NaCl

Other Types of Reactions  Combustion  Neutralization  Endothermic/ Exothermic  Oxidation/ Reduction  Precipitation

Combustion Combustion is a vigorous and exothermic reaction that takes place between certain substances (particularly organic compounds) with oxygen

Valence ➼property of an element ➼combining capacities of many elements ➼associated with the number of electrons in the outermost principal quantum level of an atom

Radicals Groups of atoms that remains intact throughout many chemical reactions and which exhibit constant valence in all compounds in which it is present.

Stoichiometry ➼Greek for “measuring elements” ➼The calculations of quantities in chemical reactions based on a balanced equation. ➼We can interpret balanced

In terms of Particles ➼Element - atoms ➼Molecular compound (non-metals) - molecule ➼Ionic Compounds (metal and non-metal) - formula unit

2H2 + O2 → 2H2O ➼Two molecules of hydrogen and one molecule of oxygen form two molecules of water. ➼2 Al2O3 → 4Al + 3O2 2formula unitsAl2O3 form 4atoms Al and 3molecules O2 2Na + 2H2O → 2NaOH + H2

Look at it differently ➼2H2 + O2 → 2H2O ➼2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. ➼2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water. ➼2 moles of hydrogen and 1 mole of

In terms of Moles ➼2 Al2O3 → 4Al + 3O2 ➼2Na + 2H2O → 2NaOH + H2 ➼The coefficients tell us how many moles of each kind

In terms of mass ➼The law of conservation of mass applies ➼We can check using moles

➼2H2 + O2 → 2H2O 2.02 g H2

= 4.04 g H2 2 moles H2 1 moles H2 32.00 g O2

= 32.00 g O2 1 moles O2 1 moles O2 36.04 g H2 + O2 36.04 H2O

In terms of mass 2H2 + O2 → 2H2O 18.02 g H2O =36.04 g H2O 2 moles H2O 1 mole H2O

2H2 + O2 → 2H2O 36.04 g (2H2 + O2) = 36.04 g H2O

Your turn Show that the following equation follows the Law of conservation of mass. 2 Al2O3 → 4Al + 3O2

Mole to Mole conversions ➼How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? ➼2 Al2O3 → 4Al + 3O2 3.34 moles Al2O3

3 mole O2 2 moles Al2O3

= 5.01 moles O2

Your turn 2C2H2 + 5 O2 → 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? How many moles of C2H2 are needed to produce 8.95 mole of H O?

Mass in Chemical Reactions

For example... ✓ If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? ✓ Fe + CuSO4 → Fe2(SO4)3 + Cu ✓ 2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu 10.1 g Fe

1 mol Fe 55.85 g Fe

= 0.181 mol Fe

2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu

3 mol Cu 0.181 mol Fe = 0.2715 mol Cu 2 mol Fe 63.55 g Cu 0.2715 mol Cu = 17.25 g Cu 1 mol Cu

Another way of doing it ... 10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu 55.85 g Fe 2 mol Fe 1 mol Cu = 17.24 g Cu

Examples To make silicon for computer chips they use this reaction SiCl4 + 2Mg → 2MgCl2 + Si How many grams of Mg are needed to make 9.3 g of Si? How many grams of SiCl4 are needed to make 9.3 g of Si? How many grams of MgCl2 are produced along with 9.3 g of

More Examples ✓The U. S. Space Shuttle boosters use this reaction ✓3Al(s) + 3NH4ClO4 → Al2O3 + AlCl3 + 3NO + 6H2O ✓How much Al must be used to react with 652 g of NH4ClO4 ? ✓How much water is produced? ✓How much AlCl3 is produced?

And still another example If 6.45 moles of water are decomposed, how many liters of oxygen gas will be produced at STP?

To compute it... ✓ If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? ✓ 2H2O → 2H2 + O2 6.45 g H2O 1 mol H2O

1 mol O2 22.42 L O2 18.02 g H2O 2 mol H2O 1 mol O2 4.01 L O2

Your Turn How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ? What volume of oxygen will be required?

Did you get it 2C4H10 + 13O2 → 8CO2 + 10H2O 23.2 g C4H10

1 mol C4H10 58.1 g C4H10 22.42 L CO2 1 mol CO2 35.81 L CO2

8 mol CO2 2 mol C4H10

Did they got it 2C4H10 + 13O2 → 8CO2 + 10H2O 23.2 g C4O10

1 mol C4H10 58.1 g C4H10 22.42 L O2 1 mol O2 58.19 L O2

13 mol O2 2 mol C4H10

Gases and Reactions

Quiz ➼How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? ➼CH4 + 2O2 → CO2 + 2H2O 1 mol O2 1 mol CH4 22.4 L CH4 17.5 L O2 22.4 L O2 2 mol O2 1 mol CH4 = 8.75 L CH4

Avogadro told us ➼Equal volumes of gas, at the same temperature and pressure contain the same number of particles. ➼Moles are numbers of particles ➼You can treat reactions as if they happen liters at a time,

Example ➼How many liters of CO2 at STP are produced by completely burning 17.5 L of CH4 ? ➼CH4 + 2O2 → CO2 + 2H2O 17.5 L CH4

1 L CO2 1 L CH4

= 17.5 L CO2

Limiting Reagent ➼If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make. ➼The limiting reagent is the reactant you run out of first. ➼The excess reagent is the one you have left over. ➼The limiting reagent determines how

How do you find out? ➼Do two stoichiometry problems. ➼The one that makes the least product is the limiting reagent. ➼For example ➼Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

➼If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? Cu is ➼2Cu + S → Cu2S Limiting 1 mol Cu2S 159.16 g Cu2S 1 mol Cu Reagent 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 13.3 g Cu2S 3.83 g S

1 mol S 32.06g S

1 mol Cu2S 159.16 g Cu2S 1 mol S 1 mol Cu2S

= 19.0 g Cu2S

Your turn ➼If 10.1 g of magnesium and 2.87 L of HCl gas are reacted at STP, how many liters of gas will be produced? .128

➼Identify the limiting reagent. ➼How many grams of solid is produced? 6.1

➼How much excess reagent remains? 8.544

Your Turn Again ➼If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced? ➼How much excess reagent will remain?

Yield ➼The amount of product made in a chemical reaction. ➼There are three types ➼Actual yield- what you get in the lab when the chemicals are mixed ➼Theoretical yield- what the balanced equation tells you you should make. ➼Percent yield = Actual x 100 % Theoretical

Example ➼6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. ➼2Al + 3 CuSO4 → Al2(SO4)3 + 3Cu ➼What is the actual yield? ➼What is the theoretical yield? ➼What is the percent yield?

13.85

49

Chemical Thermodynamics Energy in Chemical Reactions How Much? In or Out?

Energy ➼Energy is measured in Joules or calories ➼Every reaction has an energy change associated with it ➼Exothermic reactions release energy, usually in the form of heat. ➼Endothermic reactions absorb energy

Energy

C + O2 → CO2+ 395 kJ C + O2 395kJ C + O2 Reactants

→

Products

In terms of bonds C

O O

C

O O

Breaking this bond will require energy O C O C O O Making these bonds gives you energy In this case making the bonds gives you more energy than breaking them

Exothermic ➼The products are lower in energy than the reactants ➼Releases energy

Energy

CaCO → CaO + CO CaCO + 176 kJ → CaO 3 2+ CO2 3 CaO + CO2 176 kJ CaCO3 Reactants

→

Products

Endothermic ➼The products are higher in energy than the reactants ➼Absorbs energy

Chemistry Happens in ➼MOLES ➼An equation that includes energy is called a thermochemical equation ➼CH4 + 2 O2 → CO2 + 2 H2O + 802.2 kJ ➼1 mole of CH4 makes 802.2 kJ of energy. ➼When you make 802.2 kJ you

CH4 + 2 O2 → CO2 + 2 H2O + 802.2 kJ

If 10. 3 grams of CH4 are burned completely, how much heat will be produced? 10. 3 g CH4

1 mol CH4 16.05 g CH4 =514 kJ

802.2 kJ 1 mol CH4

CH4 + 2 O2 → CO2 + 2 H2O + 802.2 kJ

✓How many liters of O2 at STP would be required to produce 23 kJ of heat? ✓How many grams of water would be produced with 506 kJ of heat?

Enthalpy ➼The heat content a substance has at a given temperature and pressure ➼Can’t be measured directly because there is no set starting point ➼The reactants start with a heat content ➼The products end up with a heat content ➼So we can measure how much

Enthalpy ➼Symbol is H ➼Change in enthalpy is ∆H ➼delta H ➼If heat is released the heat content of the products is lower ➼∆H is negative (exothermic) ➼If heat is absorbed the heat content of the products is higher ➼∆H is positive (endothermic)

Energy

Change is down ∆H is <0

Reactants →

Products

Energy

Change is up ∆H is > 0

Reactants →

Products

Heat of Reaction ➼The heat that is released or absorbed in a chemical reaction ➼Equivalent to ∆H ➼C + O2(g) → CO2(g) +393.5 kJ ➼C + O2(g) → CO2(g)

∆H = -393.5 kJ

➼In thermochemical equation it is important to say what state ➼H2(g) + 1/2O2 (g)→ H2O(g) ∆H = -241.8 kJ ➼H2(g) + 1/2O2 (g)→ H2O(l) ∆H = -285.8 kJ

Heat of Combustion ➼The heat from the reaction that completely burns 1 mole of a substance ➼C2H4 + 3 O2 → 2 CO2 + 2 H2O ➼C2H6 + O2 → CO2 + H2O ➼2 C2H6 + 5 O2 → 2 CO2 + 6 H2O ➼C2H6 + (5/2) O2 → CO2 + 3 H2O

Standard Heat of Formation ➼The ∆H for a reaction that produces 1 mol of a compound from its elements at standard conditions 0 ➼Standard conditions: 25°C and 1 ∆H f atm. ◆The standard heat of formation of an ➼element Symbolisis0 ◆This

includes the diatomics

What good are they? ➼There are tables of heats of formations ➼The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products ∆H = ∆H 0f (products ) - ∆H 0f (reactants)

Examples • CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

0 ∆H f 0 ∆H f 0 ∆H f 0 ∆H f

CH4 (g) = -74.86 kJ O2(g) = 0 kJ CO2(g) = -393.5 kJ

H2O(g) = -241.8 kJ ◆ ∆H= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)] ◆ ∆H= 802.4 kJ

Why Does It Work? ✓ If H2(g) + 1/2 O2(g)→ H2O(g) ∆H=-285.5 kJ ✓ then: H2O(g) → H2(g) + 1/2 O2(g) ∆H =+285.5 kJ ➼If you turn an equation around, you change the sign ✓ 2 H2O(g) → 2 H2(g) + O2(g) ∆H =+571.0 kJ ➼If you multiply the equation by a number, you multiply the heat by that number.

Why does it work? ➼You make the products, so you need their heats of formation ➼You “unmake” the products so you have to subtract their heats.

Practice Problem Diborane, B2H6 is a highly reactive boron hydride. Calculate the ∆H for the synthesis of diborane from its elements according to the equation 2 B (s) + 3 H2 (g) → B2H6 (g) and the following data: 2B(s) + 3/2O2(g) → B2O3 -1273 kJ B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) -2035 kJ

2B

(s)

+ 3 H2 (g) → B2H6 (g)

2B(s) + 3/2O2(g) → B2O3 -1273 kJ B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) -2035 kJ H2(g) + 1/2O2(g) → H2O(l) -286 kJ

Hess’s Law In going from a particular set of reactants to a particular set of products, the enthalpy change is the same whether the reaction takes place in one step or in a series of steps

Quiz ✓Find the enthalpy change for the chemical reaction given below: 2 SO3(g) → 2 SO2(g) + O2(g) 0 ∆H f

SO2 (g)

-297 kJ/mol

SO3 (g)

-396 kJ/mol

0 ∆H f

SO2 (g)

-297 kJ/mol

SO3 (g)

-396 kJ/mol

2(-297) - [ 2(-396)] = 198

198 kJ/mol

1st Law of Thermodynamics

Energy can neither be created nor destroyed the capacity to do work or cause heat flow

Types of Energy ➼Potential Energy - energy due to position or composition ➼Kinetic Energy - energy due to motion and depends on mass and velocity of the

nd Law of Thermodynamics

In any spontaneous process there is always an increase in the entropy of the universe the entropy of the universe is increasing

Entropy ➼A thermodynamic function ➼Symbol is S ➼It is a measure of randomness or disorder ➼A driving force for spontaneous process

Entropy (S) ➼Change in entropy is ∆S (delta S) ➼If the reaction is accompanied by production of heat ∆S is positive ➼If the reaction involves heat absorption ∆S is negative ➼If the reaction neither

Change in Entropy (∆S) • ∆S > 0 spontaneous reaction • ∆S < 0 not spontaneous reaction • ∆S = 0 system at equilibrium

Factors Affecting ∆S

➼ ➼

Heat flow Temperature

∆Suniv = ∆Ssys + ∆Ssurr Consider liquid water, H2O H2 O

(l)

→ H2 O

(g)

(l)

44 kJ

∆Ssys ∆Ssurr ∆suniv

spontaneous?

+

+

+

Yes

–

–

–

No

+

–

∆Ssys>∆Ssurr

Yes

+

–

∆Ssys<∆Ssurr

No

–

+

∆Ssys>∆Ssurr

No

–

+

∆Ssys<∆Ssurr

Yes

What have you learned?

Ssurr is temperature dependen

∆Ssurr

∆H = T

Practice Problem Calculate the ∆Ssurr for the evaporation of water at 25 °C and at 100 °C. .15.12

H2 O

(l)

→ H2 O

(g)

44 kJ

Gibbs Free Energy

Free Energy ➼Symbol is G (in honor of Josiah Willard Gibbs) ➼Another thermodynamic function ➼related to the spontaneity of a reaction or process ➼useful in dealing with temperature dependence of spontaneity ➼defined as G = H - TS

Change in Free Energy (∆ G) Change in free energy is ∆G (delta G)

∆G = ∆H - T∆S ∆G° = ∆H° - T∆S°

∆Suniv =

∆G T

Note: A process at constant T and P is spontaneous in the direction in which the free

Case ∆S positive, ∆H negaitive ∆S positive, ∆H positive ∆S negative, ∆H negative ∆S negative, ∆H positive

Result Spontaneous at all temperature Spontaneous at high temperatures Spontaneous at low temperature NOT spontaneous at any temperature

Br2 (l) → Br2 (g) At what temperature is the process spontaneous at 1 atm? ∆H° = 31.0 kJ/mol ∆S° = 93.0 ∆G° = ∆H° - T∆S° = 0 J/K•mol → T = ∆H°/∆S° →

= 333 °Κ

Ans. = T > 333 °K or T > 60

H2O (s) → H2O (l) At what temperature is the process spontaneous at 1 atm? ∆H° = 6033 J/mol ∆S° = 22.1 ∆G° = ∆H° - T∆S° = 0 J/K•mol → T = ∆H°/∆S° →

= 273 °Κ

Ans. = Τ > 0 °C

H2O (l) → H2O (g) At what temperature is the process spontaneous at 1 atm? ∆H° = 44.4 kJ/mol ∆S° = 119 ∆G° = ∆H° - T∆S° = 0 J/K•mol → T = ∆H°/∆S° →

= 373 °Κ

Ans. = Τ > 100 °C

Chemical Kinetics

Chemical Reaction ➼Can be classified into several types like neutralization, precipitation, redox, etc. ➼Affected by several factors like nature of reactants, pressure, temperature, etc. ➼It is defined or represented by its reactants and products

Chemical Kinetics The area of chemistry that is involved in rates of chemical reactions

Reaction Rates? defined as the change in concentration of reactant or product per unit time

∆[Α] Rate = ∆t

2 NO2 (g) → 2 NO

(g)

+ O2 (g)

Concentration (mol/L) Time (±1 s)

NO2

NO

O2

0 50

0.0100 0.0079

0.0000 0.0021

0.0000 0.0011

100

0.0065

0.0035

0.0018

150

0.0055

0.0045

0.0023

Calculate the average rate at which the concentration of NO2 changes over the first 50 ∆[ΝΟ2] seconds. ∆[Α] Rate = =

∆t

=

∆t

[ΝΟ2]t = 50 s - [ΝΟ2]t = 0 s 50 s - 0 s

= - 4.2 x 10-5 M/s

Collision model ➼ A model based on the idea that atoms/molecules must collide to react ➼ The model used to account for the observed behavior and characteristics of reaction rates.

Energy The threshold energy that must be overcome to produce a chemical reaction

2 BrNO (g) → 2 NO

(g)

+ Br2 (g)

➼ 2 Br-N bonds must be broken ➼ 1 Br-Br bond must be formed ➼ the kinetic energy possessed by the reacting molecules, BrNO + BrNO is required to break the bond

Potential Energy

ON---Br ON---Br

bond breaking bond forming

Ea 2 BrNO

}∆E

2NO + Br2

Reaction Progress

Complex or Transition State

The arrangement of atoms found at the top of the potential energy barrier as reaction proceeds to completion